# Thread: A basic function limit question

1. ## A basic function limit question

We've just started doing continuity, limits of functions etc, and I'm not quite sure how to progress.

If we suppose g and h are real-valued functions, defined on some interval (m,n) containing $
x_0
$
, and $
\mathop {\lim }\limits_{x \to x_0 } g\left( x \right) = k
$
and $
\mathop {\lim }\limits_{x \to x_0 } h\left( x \right) = l
$
.

I want to show that if $
g\left( x \right) < h\left( x \right){\text{ }}\forall x \in \left( {x_0 - \delta ,x_0 + \delta } \right),\delta > 0
$
then $
k \leqslant l
$

Also, do you know of any good resources to get further help on this topic? I know the members here are extremely knowledgeable, but do you have any online text resource etc that may help? Maybe with examples of a similar kind?

2. Originally Posted by StandardToaster
We've just started doing continuity, limits of functions etc, and I'm not quite sure how to progress.

If we suppose g and h are real-valued functions, defined on some interval (m,n) containing $
x_0
$
, and $
\mathop {\lim }\limits_{x \to x_0 } g\left( x \right) = k
$
and $
\mathop {\lim }\limits_{x \to x_0 } h\left( x \right) = l
$
.

I want to show that if $
g\left( x \right) < h\left( x \right){\text{ }}\forall x \in \left( {x_0 - \delta ,x_0 + \delta } \right),\delta > 0
$
then $
k \leqslant l
$

suppose $k > l$ and choose $\epsilon = \frac{k-l}{2}.$ then by the definition of limit, there exists $\delta_1 > 0$ such that $g(x) > k - \epsilon$ if $|x-x_0| < \delta_1.$ also there exists $\delta_2 > 0$ such that $h(x) < l + \epsilon$ if $|x-x_0| < \delta_2.$

now let $\delta_0=\min \{\delta, \delta_1, \delta_2\}.$ so if $|x-x_0| < \delta_0,$ then we'll have: $l + \epsilon > h(x) > g(x) > k- \epsilon,$ and thus: $\epsilon > \frac{k-l}{2}.$ contradiction!